Question

What is the empirical formula for a compound comprised of 1.8% hydrogen, 56.1% sulfur and 42.1% oxygen?

Answer 1

Hope this Will help you

Please mark as brainlist.......

Answer 2

The empirical formula for a compound is $H_2S_2O_3$

Explanation:

We need to find the empirical formula for a compound comprised of 1.8% hydrogen, 56.1% sulfur and 42.1% oxygen.

Firstly, finding the moles of each element as :

Hydrogen :

$1.8\ g\times \dfrac{1\ mol}{1.008\ g}=1.78\ mol\ H$

Sulfur :

$56.1\ g\times \dfrac{1\ mol}{32.06\ g}=1.749\ mol\ S$

Oxygen :

$42.1\ g\times \dfrac{1\ mol}{15.999\ g}=2.63\ mol\ O$

Now taking mole ratio :

Hydrogen : $\dfrac{1.78}{1.78}=1$

Sulfur : $\dfrac{1.749}{1.78}=0.98$

Oxygen : $\dfrac{2.63}{1.78}=1.47$

Multiply each ratio by 2 so that we get whole number. So,

H = 2

S = 1.96 or 2

O = 2.95 or 3

So, the empirical formula for a compound is $H_2S_2O_3$. Hence, this is the required solution.

Learn more,

Empirical formula

https://brainly.in/question/1360363